Báo cáo hóa học: " Research Article Eigenvalue Problems for Systems of Nonlinear Boundary Value Problems on Time Scales" pptx

On a larger scale, there has been a great deal of study focused on positive solutions of boundary value problems for ordinary differential equations.. He, “Double positive solutions of bo

Volume 2007, Article ID 31640, 10 pages

doi:10.1155/2007/31640

Research Article

Eigenvalue Problems for Systems of Nonlinear Boundary Value Problems on Time Scales

M Benchohra, J Henderson, and S K Ntouyas

Received 28 June 2007; Accepted 19 November 2007

Recommended by Kanishka Perera

Values ofλ are determined for which there exist positive solutions of the system of

dy-namic equations, uΔΔ(t) + λa(t) f (v(σ(t))) =0, vΔΔ(t) + λb(t)g(u(σ(t))) =0, for t ∈

[0, 1]T, satisfying the boundary conditions, u(0) =0= u(σ2(1)),v(0) =0= v(σ2(1)),

where T is a time scale A Guo-Krasnosel’skii fixed point-theorem is applied.

Copyright © 2007 M Benchohra et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Let T be a time scale with 0,σ2(1)∈T Given an intervalJ ofR, we will use the interval notation

We are concerned with determining values of λ (eigenvalues) for which there exist

positive solutions for the system of dynamic equations

uΔΔ(t) + λa(t) fvσ(t)=0, t ∈[0, 1]T,

vΔΔ(t) + λb(t)guσ(t)=0, t ∈[0, 1]T, (1.2) satisfying the boundary conditions

u(0) =0= uσ2(1)

, v(0) =0= vσ2(1)

(a) f ,g ∈ C([0,∞), [0,∞)),

(b)a,b ∈ C([0,σ(1)]T, [0,∞)), and each does not vanish identically on any closed subinterval of [0,σ(1)]T,

(c) all of f0:=limx →0 +(f (x)/x), g0:=limx →0 +(g(x)/x), f ∞:=limx →∞(f (x)/x), and

g ∞:=limx →∞( g(x)/x) exist as real numbers.

There is an ongoing flurry of research activities devoted to positive solutions of dy-namic equations on time scales (see, e.g., [1–7]) This work entails an extension of the paper by Chyan and Henderson [8] to eigenvalue problems for systems of nonlinear boundary value problems on time scales Also, in that light, this paper is closely related

to the works of Li and Sun [9,10]

On a larger scale, there has been a great deal of study focused on positive solutions of boundary value problems for ordinary differential equations Interest in such solutions is high from a theoretical sense [11–15] and as applications for which only positive solutions are meaningful [16–19] These considerations are caste primarily for scalar problems, but good attention has been given to boundary value problems for systems of differential equations [20–24]

The main tool in this paper is an application of the Guo-Krasnosel’skii fixed point-theorem for operators leaving a Banach space cone invariant [12] A Green function plays

a fundamental role in defining an appropriate operator on a suitable cone

2 Some preliminaries

In this section, we state the well-known Guo-Krasnosel’skii fixed point-theorem which

we will apply to a completely continuous operator whose kernel,G(t,s), is the Green

function for

−yΔΔ=0,

y(0) =0= yσ2(1)

Erbe and Peterson [6] have found that

G(t,s) = σ21(1)

⎧

⎨

⎩

tσ2(1)− σ(s), ift ≤ s, σ(s)σ2(1)− t, ifσ(s) ≤ t, (2.2)

from which

G(t,s) > 0, (t,s) ∈0,σ2(1)

G(t,s) ≤ Gσ(s),s= σ(s)σ2(1)− σ(s)

σ2(1) , t ∈0,σ2(1)

T,s ∈0,σ(1)T, (2.4) and it is also shown in [6] that

G(t,s) ≥ kGσ(s),s= k σ(s)



σ2(1)− σ(s)

σ2(1) , t ∈ σ2(1)

3σ2(1)

4 T,s ∈0,σ(1)T,

(2.5)

k =min

 1

4,

σ2(1)

4

σ2(1)− σ(0)

We note that a pair (u(t),v(t)) is a solution of the eigenvalue problem (1.2), (1.3) if and only if

u(t) = λ σ(1)

0 G(t,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs, 0 ≤ t ≤ σ2(1),

v(t) = λ σ(1)

0 G(t,s)b(s)guσ(s)Δs, 0 ≤ t ≤ σ2(1).

(2.7) Values ofλ for which there are positive solutions (positive with respect to a cone) of

(1.2), (1.3) will be determined via applications of the following fixed point-theorem [12]

Theorem 2.1 Let Ꮾ be a Banach space, and let ᏼ ⊂ Ꮾ be a cone in Ꮾ Assume that Ω1

andΩ2are open subsets of Ꮾ with 0 ∈Ω1⊂Ω1⊂Ω2, and let

T : ᏼ ∩Ω2\Ω1



be a completely continuous operator such that either

(i) Tu ≤ u , u ∈ᏼ∩ ∂Ω1, and Tu ≥ u , u ∈ᏼ∩ ∂Ω2, or

(ii) Tu ≥ u , u ∈ᏼ∩ ∂Ω1, and Tu ≤ u , u ∈ᏼ∩ ∂Ω2.

Then, T has a fixed point in ᏼ ∩(Ω2\Ω1).

3 Positive solutions in a cone

In this section, we applyTheorem 2.1to obtain solutions in a cone (i.e., positive solu-tions) of (1.2), (1.3) Assume throughout that [0,σ2(1)]Tis such that

ξ =min

t ∈ T | t ≥ σ2(1)

4

 ,

ω =max

t ∈ T | t ≤3σ2(1)

4

;

(3.1)

both exist and satisfy

σ2(1)

4 ≤ ξ < ω ≤3σ2(1)

Next, letτ ∈[ξ,ω]Tbe defined by

ω

ξ G(τ,s)a(s)Δs = max

t ∈[ξ,ω]T

ω

Finally, we define

l = min

s ∈[0,σ2 (1)]

Gσ(ω),s

and let

For our construction, let Ꮾ= {x : [0,σ2(1)]T→R} with supremum norm x =

sup{|x(t)|:t ∈[0,σ2(1)]T}and define a coneᏼ⊂Ꮾ by

ᏼ=x ∈Ꮾ| x(t) ≥0 on

0,σ2(1)

T, andx(t) ≥ m x , fort ∈ξ,σ(ω)T. (3.6) For our first result, define positive numbersL1andL2by

L1:=max

m ω

ξ G(τ,s)a(s)Δs f ∞

−1 ,

m ω

ξ G(τ,s)b(s)Δsg ∞

−1 ,

L2:=min

σ(1)

0 Gσ(s),sa(s)Δs f0

−1 ,

σ(1)

0 Gσ(s),sb(s)Δsg0

−1 , (3.7)

where we recall thatG(σ(s),s) = σ(s)(σ2(1)− σ(s))/σ2(1).

Theorem 3.1 Assume that conditions (a), (b), and (c) are satisfied Then, for each λ satis-fying

there exists a pair (u,v) satisfying ( 1.2 ), ( 1.3 ) such that u(x) > 0 and v(x) > 0 on (0,σ2(1))T Proof Let λ be as in (3.8) And let > 0 be chosen such that

max

m ω

ξ G(τ,s)a(s)Δsf ∞ −  −

1 ,

m ω

ξ G(τ,s)b(s)Δsg ∞ −  −

1

≤ λ,

λ ≤min

σ(1)

0 Gσ(s),sa(s)Δsf0+

−1 ,

σ(1)

0 Gσ(s),sb(s)Δsg0+

−1

.

(3.9) Define an integral operatorT : ᏼ→Ꮾ by

Tu(t) := λ σ(1)

0 G(t,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs, u ∈ ᏼ. (3.10)

By the remarks inSection 2, we seek suitable fixed points ofT in the cone ᏼ.

Notice from (a), (b), and (2.3) that, for u ∈ ᏼ, Tu(t) ≥0 on [0,σ2(1)]T Also, for

u ∈ᏼ, we have from (2.4) that

Tu(t) = λ σ(1)

0 G(t,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≤ λ σ(1) Gσ(s),sa(s) fλ σ(1) Gσ(s),rb(r)guσ(r)ΔrΔs

(3.11)

so that

Tu ≤ λ σ(1)

0 Gσ(s),sa(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs. (3.12) Next, ifu ∈ᏼ, we have from (2.5), (3.5), and (3.10) that

min

t ∈[ξ,ω]T

Tu(t) = min

t ∈[ξ,ω]T

λ σ(1)

0 G(t,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≥ λm σ(1)

0 Gσ(s),sa(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≥ m Tu

(3.13) Consequently,T : ᏼ→ ᏼ In addition, standard arguments show that T is completely

con-tinuous

Now, from the definitions of f0andg0, there existsH1> 0 such that

f (x) ≤f0+x, g(x) ≤g0+x, 0< x ≤ H1. (3.14) Letu ∈ᏼ with u = H1 We first have from (2.4) and choice of, for 0≤ s ≤ σ(1),

that

λ σ(1)

0 Gσ(s),rb(r)guσ(r)Δr ≤ λ σ(1)

0 Gσ(r),rb(r)guσ(r)Δr

≤ λ σ(1)

0 Gσ(r),rb(r)g0+u(r)Δr

≤ λ σ(1)

0 Gσ(r),rb(r)Δrg0+ u

≤ u = H1.

(3.15)

As a consequence, we next have from (2.4) and choice of, for 0≤ t ≤ σ2(1), that

Tu(t) = λ σ(1)

0 G(t,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≤ λ σ(1)

0 Gσ(s),sa(s)f0+λ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≤ λ σ(1)

0 Gσ(s),sa(s)f0+H1Δs

≤ H1= u

(3.16)

So, Tu ≤ u If we set

Ω1=x ∈Ꮾ | x < H1 

then

Tu ≤ u , foru ∈ᏼ∩ ∂Ω1. (3.18)

Next, from the definitions of f ∞andg ∞, there existsH2> 0 such that

f (x) ≥f ∞ − x, g(x) ≥g ∞ − x, x ≥ H2. (3.19) Let

H2=max

2H1,H2

m

Letu ∈ᏼ and u = H2 Then,

min

t ∈[ξ,ω]T

Consequently, from (2.5) and choice of, for 0≤ s ≤ σ(1), we have that

λ σ(1)

0 Gσ(s),rb(r)guσ(r)Δr ≥ λ ω

ξ Gσ(s),rb(r)guσ(r)Δr

≥ λ ω

ξ G(τ,r)b(r)guσ(r)Δr

≥ λ ω

ξ G(τ,r)b(r)g ∞ − u(r)Δr

≥ mλ ω

ξ G(τ,r)b(r)g ∞ − Δr u

≥ u = H2.

(3.22)

And so, we have from (2.5) and choice ofthat

Tu(τ) = λ σ(1)

0 G(τ,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≥ λ σ(1)

0 G(τ,s)a(s)f ∞ − λ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≥ λ σ(1)

0 G(τ,s)a(s)f ∞ − H2Δs

≥ mH2> H2= u

(3.23)

Hence, Tu ≥ u So, if we set

Ω2=x ∈Ꮾ | x < H2



then

Tu ≥ u , foru ∈ᏼ∩ ∂Ω2. (3.25)

ApplyingTheorem 2.1to (3.18) and (3.25), we obtain thatT has a fixed point u ∈

ᏼ∩(Ω2\Ω1) As such, and withv being defined by

v(t) = λ σ(1)

0 G(t,s)b(s)guσ(s)Δs, (3.26) the pair (u,v) is a desired solution of (1.2), (1.3) for the givenλ The proof is complete.

 Prior to our next result, we introduce another hypothesis

(d)g(0) =0, andf is an increasing function.

We now define positive numbersL3andL4by

L3:=max

m ω

ξ G(τ,s)a(s)Δs f0

−1 ,

m ω

ξ G(τ,s)b(s)Δsg0

−1 ,

L4:=min

σ(1)

0 Gσs(s)a(s)Δs f ∞

−1 ,

σ(1)

0 Gσs(s)b(s)Δsg ∞

−1

.

(3.27)

Theorem 3.2 Assume that conditions (a)–(d) are satisfied Then, for each λ satisfying

there exists a pair (u,v) satisfying ( 1.2 ), ( 1.3 ) such that u(x) > 0 and v(x) > 0 on (0,σ2(1))T Proof Let λ be as in (3.28) And let > 0 be chosen such that

max

m ω

ξ G(τ,s)a(s)Δsf0−  −

1 ,

m ω

ξ G(τ,s)b(s)Δsg0−  −

1

≤ λ,

λ ≤min

σ(1)

0 Gσ(s),sa(s)Δsf ∞+

−1 ,

σ(1)

0 Gσ(s),sb(s)Δsg ∞+

−1

.

(3.29) Let T be the cone preserving, completely continuous operator that was defined by

(3.10)

From the definitions of f0andg0, there existsH1> 0 such that

f (x) ≥f0− x, g(x) ≥g0− x, 0< x ≤ H1. (3.30) Now,g(0) =0, and so there exists 0< H2< H1such that

σ(1)

0 Gσ(s),sb(s)Δs, 0≤ x ≤ H2. (3.31)

Chooseu ∈ᏼ with u = H2 Then, for 0≤ s ≤ σ(1), we have

λ0σ(1) Gσ(s),rb(r)guσ(r)Δr ≤

σ(1)

0 Gσ(s),rb(r)H1Δr

σ(1)

Gσ(s),sb(s)Δs ≤ H1. (3.32)

Tu(τ) = λ σ(1)

0 G(τ,s)a(s) fλ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≥ λ ω

ξ G(τ,s)a(s)f0− λ σ(1)

0 Gσ(s),rb(r)guσ(r)ΔrΔs

≥ λ ω

ξ G(τ,s)a(s)f0− λ ω

ξ G(τ,r)b(r)guσ(r)ΔrΔs

≥ λ ω

ξ G(τ,s)a(s)f0− λm ω

ξ G(τ,r)b(r)g0−  u ΔrΔs

≥ λ ω

ξ G(τ,s)a(s)f0−  u Δs

≥ λm ω

ξ G(τ,s)a(s)f0−  u Δs ≥ u

(3.33)

So, Tu ≥ u If we put

Ω1=x ∈Ꮾ | x < H2



then

Tu ≥ u , foru ∈ᏼ∩ ∂Ω1. (3.35) Next, by definitions of f ∞andg ∞, there existsH1such that

f (x) ≤f0− x, g(x) ≤g0− x, x ≥ H1. (3.36) There are two cases: (a)g is bounded, and (b) g is unbounded.

For case (a), supposeN > 0 is such that g(x) ≤ N for all 0 < x < ∞ Then, for 0≤ s ≤ σ(1) and u ∈ᏼ,

λ σ(1)

0 Gσs(r)b(r)guσ(r)Δr ≤ Nλ σ(1)

0 Gσ(r),rb(r)Δr. (3.37) Let

M =max

f (x) |0≤ x ≤ Nλ σ(1)

0 Gσ(r),rb(r)Δr, (3.38) and let

H3> max2H2,Mλ σ(1)

0 Gσ(s),sa(s)Δs. (3.39) Then, foru ∈ᏼ with u = H3,

Tu(t) ≤ λ σ(1)

0 Gσ(s),sa(s)MΔs

so that Tu ≤ u If

Ω2=x ∈Ꮾ | x < H3



then

Tu ≤ u , foru ∈ᏼ∩ ∂Ω2. (3.42) For case (b), there exists H3> max{2H2,H1} such that g(x) ≤ g(H3), for 0< x ≤

H3 Similarly, there exists H4> max {H3,λσ(1)0 G(σ(r),r)b(r)g(H3)Δr) }such that f (x) ≤

f (H4), for 0< x ≤ H4 Choosing u ∈ᏼ with u = H4we have by (d) that

Tu(t) ≤ λ σ(1)

0 G(t,s)a(s) fλ σ(1)

0 Gσ(r),rb(r)gH3



ΔrΔs

≤ λ σ(1)

0 G(t,s)a(s) fH4



Δs

≤ λ σ(1)

0 Gσ(s),sa(s)Δsf ∞+H4

≤ H4= u ,

(3.43)

and so Tu ≤ u For this case, if we let

Ω2=x ∈Ꮾ | x < H4



then

Tu ≤ u , foru ∈ᏼ∩ ∂Ω2. (3.45)

In either cases, application of part (ii) ofTheorem 2.1yields a fixed pointu of T

be-longing toᏼ∩(Ω2\Ω1), which in turn yields a pair (u,v) satisfying (1.2), (1.3) for the

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M Benchohra: Laboratoire de Math´ematiques, Universit´e de Sidi Bel Abb`es, BP 89,

22000 Sidi Bel Abb`es, Algeria

Email address:benchohra@univ-sba.dz

J Henderson: Department of Mathematics, Baylor University, Waco, TX 76798, USA

Email address:johnny henderson@baylor.edu

S K Ntouyas: Department of Mathematics, University of Ioannina, 45110 Ioannina, Greece

Email address:sntouyas@uoi.gr

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